Introduction - If you have any usage issues, please Google them yourself
If we dye 12 sides with the color of k and have no limits on the number of colors per color (the previous zzwu tries to solve the problem), then we can use the Polya theorem in the group theory. So if you just take all of the x1's, x2,x3, x4, you're going to replace it with k, and then you're going to divide the result by 24, which is to say, you're going to have 12 edges of the color in k
12 + 6 (k ^ ^ 7 + 3 * * k k k ^ ^ 6 + 8 * 4 + 6 * k ^ 3) / 24 kinds
But for the problem in this problem, not only the total number of colors used, but also the number of each color used, which is a lot more complicated and can be used
Cauchy - Frobenius - Burnside lemma.
The number of colors is equal to
{Sum {|Fix(P)|}/|G|, for each P in G}
G is this group with 24 elements.
Fix(P) refers to the number of dyeing schemes that satisfy the following conditions:
After replacing P with the side of this dyeing scheme, the scheme of dyeing is still the scheme (no transformation).
That is, for the dyeing scheme in Fix (P), all the edges in one ring are stained with the same color.
And then we just have to do the calculation for P divided by 5.